Suppose, I have a $3 \times 5$ matrix of the form,
$A=\left(
\begin{array}{ccccc}
a_{1,1} & a_{1,2} & a_{1,3} & a_{1,4} & a_{1,5} \\
a_{2,1} & a_{2,2} & a_{2,3} & a_{2,4} & a_{2,5} \\
a_{3,1} & a_{3,2} & a_{3,3} & a_{3,4} & a_{3,5} \\
\end{array}
\right)=\left( a_1 \, a_2 \, a_3 \, a_4 \, a_5 \right)$
where $a_i$ denotes a column matrix, and A is of rank 3.
Now, let me denote $A_{i,j,k}=( a_i \, a_j \, a_k)$.
Then, how can I prove the following identity?
$$\left|
\begin{array}{ccccc}
|A_{1,4,5}| & |A_{2,4,5}| \\
|A_{3,1,5}| & |A_{3,2,5}| \\
\end{array}
\right|=\left| A_{3,4,5} \right| \left| A_{1,2,5} \right|$$
Using Mathematica, I proved the above identity by expanding each determinant term explicitly. But I want an algebraic way to prove the above identity because I want to generalise the above identity for a general $A$ matrix of type $m\times n$.
If you can give me some references where such determinant identities are discussed, it would be great.
Best Answer
Let's assume you are working over $\mathbb{R}$. Then this follows from the Lagrange's identity on $\mathbb{R}^3$: $$ (\mathbf{A}\times\mathbf{B})\cdot(\mathbf{C}\times\mathbf{D})=(\mathbf{A}\cdot\mathbf{C})(\mathbf{B}\cdot\mathbf{D})-(\mathbf{A}\cdot\mathbf{D})(\mathbf{B}\cdot\mathbf{C}) $$ WLOG we can assume $a_5$ is a unit vector and projecting away the $a_5$-component of $a_1,\dots,a_4$ does not changing the $\det A_{i,j,5}$. In this case, $a_i\times a_j\parallel a_5$ ($1\leq i,j\leq 4$) and so you have $$ (\det A_{i,j,5})(\det A_{k,l,5})=(a_i\times a_j)\cdot(a_k\times a_l) $$ where again $1\leq i,j,k,l\leq 4$. A suitable sum then gives the result.
Generalisations other than $$ \det\begin{pmatrix} \det A_{1,4,5,\dots,n} & \det A_{2,4,5,\dots,n}\\ \det A_{3,1,5,\dots,n} & \det A_{2,3,5,\dots,n} \end{pmatrix} =(\det A_{1,2,5,\dots,n})(\det A_{3,4,5,\dots,n}) $$ for $(n-2)\times n$ matrix $A$ (similar notation as your question) may be possible using the Binet-Cauchy identity.